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Let $G$ be an Abelian group of order $mn$ where $\gcd(m,n)=1$.

Assume that $G$ contains an element of $a$ of order $m$ and an element $b$ of order $n$.

Prove $G$ is cyclic with generator $ab$.

The idea is that $(ab)^k$ for $k \in [0, \dots , mn-1]$ will make distinct elements but do not know how to argue it.

Could I say something like $<a>=A$, $<b>=B$, somehow $AB=\{ ab : a \in A , b \in B \}$ and that has order $|A||B|=mn$?

Don't know if it's the same exact or similar to Finite group of order $mn$ with $m,n$ coprime.

The Phenotype
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Tiger Blood
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3 Answers3

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Hint: (1) Part of the definition of $l=\mathrm{lcm}(m,n)$ is that if $m|k$ and $n|k$, then $l|k$. (2) if $\gcd(m,n)=1$ then $\mathrm{lcm}(m,n)=mn$.

So, if $(ab)^k=a^kb^k=1$, then $a^k=1$ and $b^k=1$. What can you conclude about $k$?

David Hill
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Let $k\in \mathbb{N}$ and assume $(ab)^k = e$. Then, since $G$ is abelian, $a^k = b^{-k}$. Raising this to the power $n$, we get $a^{nk} = e$, so $m| nk$. But $m,n$ are coprime, so by Gauss's theorem, $m|k$. Witha similar argument, $n|k$, and since again $m,n$ are coprime, we get $nm |k$. Conversely, we obviously have $(ab)^{nm}= e$ (again, it's necessary to assume $G$ to be abelian here). So $ab$ is of order $nm$, which gives $G= \langle ab \rangle$.

Maxime Ramzi
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Suppose $(ab)^k=e$. This means $a^k=b^{-k}\in\langle a\rangle\cap\langle b\rangle$. But since $m\wedge n=1$,$\;\langle a\rangle\cap\langle b\rangle=\{e\}$ by Lagrange's theorem. Thus $a^k=e$ and $b^k=e$, which implies $k $ is a common multiple of $m$ and $n$, i.e. a multiple of $mn$ since $m\wedge n=1$.

Bernard
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